Dual vector space - Lagrange Interpolating Polynomial

[SqueeSpleen]

I think I solved it a week ago, but I didn't write down all the things and I want to be sure of doing the things right, plus the excersise of writing it here in latex helps me a loot (I wrote about 3 threads and didn't submited it because writing it here clarified me enough to find the answer for my self xD). So feel free to correct me.

Lets be \alpha _{0},...,\alpha _{n} \in K, \alpha _{i} \neq \alpha _{j} if i\neq j

To every i, 0 \leq i \leq n, we define \varepsilon _{\alpha _i}:K_{n}[X]\rightarrow K as \varepsilon _{\alpha _i}(P)=P(\alpha _i).

i) Prove that B = \left \{ \epsilon _{\alpha _{0}},...,\epsilon _{\alpha _{n}} \right \} is a basis of (K_{n}[X])^{*}

ii) Let's be B = \left \{ P_{0},...,P_{n} \right \} the basis of K_{n}[X] such that B^{*}=B_{1}. Prove that the polynomial

P = \sum _{i=0}^{n} \beta _i . P_i

is the only polynomial in K[X] of grade minor or equal to n such that,

\forall i, 0 \leq i \leq n, P(\alpha _{i}) = \beta _{i}

This polynomial is named the Lagrange's Interpolating Polynomial (irrelevant :P).

iii) Prove that there're real numbers \left \a_{0},...,a_{n}{ \right \} such that, to all P \in \mathbb{R}_{n}[X]
\int _{0}^{1}P(x)dx=\sum _{i=0}^{n}a_{i}.P(\alpha _{i})
Find a_{0}, a_{1} and a_{2} when n = 2, a_{0}=1, a_{1}=\frac{1}{2} and a_{2}=0

Relevant proposition:
Lets be V a K-vector space of dimensión n, and let's be V* it's dual space.
Lets be B_{1}=\left \{ \varphi _{1},...\varphi _{n} \right \} a basis of V*. Then there's only one basis B = \left \{ v _{1},...v _{n} \right \} of V such that:
B^{*}=B_{1}

i) Suppose that:
\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}=0
Then, to all P \in K_n[X], we have: \sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}(P)=\sum _{i=0}^{n}P(\alpha _{i})=0
To every \forall i, 0 \leq i \leq n
We take the polynomial:
\prod _{i=0,{i\neq }j}^{n}(X-\alpha _{i})
And evaluating it we know \alpha _{j}=0, because P(\alpha _{i})=0 to every i≠j (P(\alpha _{j}) can't be zero because \alpha _{j} \neq \alpha _{i} if i≠j
Then, we have the linear independency of the set, and as n+1 elements, then it's generated space has dimension n+1 and it's a subset of (K_{n}[X])*, so it's a basis of (K_{n}[X])* (because it also has dimension n+1).
ii) So we know B_{1} is a basis of K_n[X], then for the proposition we know that there's only one basis
B= \left \{ P_{0},...,P_{n} \right \} such that B^{*}=B_{1}
So, by the definition of dual basis, \epsilon _{\alpha _{i}}(P_{j})=P_{j}(\epsilon _{\alpha _{0}})=
1 if i=j
0 if i≠j
Then P(\epsilon _{\alpha _{j}})=\sum _{i=0}^{n} \beta _i . P_i(\epsilon _{\alpha _{j}})=\beta _{i}

iii)
I think I'm stuck in 3, but I'll think later after sleeping, I'm probably failing due to being so sleepy.

 

[SqueeSpleen]

I didn't find the edit button in my first post, I assume someone blocked it so I bump the thread when I finish it instead of buring it in the bottom of the forum's abyss.
If P=\sum _{i=0}^{n}\beta _{i} P_{i} then
\int _{0}^{1}\sum _{i=0}^{n} \beta _{i} . P_{i} (x)dx = \sum _{i=0}^{n} \beta _{i}.\int _{0}^{1}P_{i} (x)dx
\sum _{i=0}^{n}a_i.P(\alpha _{i})=\sum _{i=0}^{n}a_i.\sum _{j=0}^{n} \beta _{i} P_{i} (\alpha _{i})=
\sum _{i=0}^{n}a_{i}.\beta _{i}
Let's be a_{i}=\int _{0}^{1}P_{i} (x)dx to all 0 \leq i \leq n
Then
\sum _{i=0}^{n} \frac{\beta _{i}}{i+1}=\sum _{i=0}^{n}a_{i}.\beta _{i}Let's do find a_{0}, a_{1} and a_{2} when n = 2, a_{0}=1, a_{1}=\frac{1}{2} and a_{2}=0
\left\{\begin{matrix}<br /> P_{0}(\alpha _{0})=1\\<br /> P_{0}(\alpha _{1})=0\\<br /> P_{0}(\alpha _{2})=0 <br /> \end{matrix}\right.
\left\{\begin{matrix}<br /> P_{0}(1)=1\\<br /> P_{0}(\frac{1}{2})=0\\<br /> P_{0}(0)=0 <br /> \end{matrix}\right.
As we know two roots of P_{0} we only need to multiply it for a constant to make it equal to 1 when evaluated in 1.
-\frac{(X-\frac{1}{2})(X)}{2}

<br /> \left\{\begin{matrix}<br /> P_{1}(1)=0\\<br /> P_{1}(\frac{1}{2})=1\\<br /> P_{1}(0)=0 <br /> \end{matrix}\right.
-4(X-1)(X)

\left\{\begin{matrix}<br /> P_{2}(1)=0\\<br /> P_{2}(\frac{1}{2})=0\\<br /> P_{2}(0)=1 <br /> \end{matrix}\right.

2(X-\frac{1}{2})(X-1)

So we only need to evaluate the integrals of these polynomials.
Then:
a_{0}=-\frac{1}{24}
a_{1}=\frac{2}{3}
a_{2}=\frac{1}{6}

PD: The excersise is from http://mate.dm.uba.ar/~jeronimo/algebra_lineal/AlgebraLineal.pdf page 115 Ejercise 16, this isn't the book used in my Linear Algebra's course in the dual vector space subject but I find the ejercises were pretty interesting to do.